Problème d'équations du second degré


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PDF Le second degré - Lycée d'Adultes

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On pose un système de deux équations à deux inconnues. Développons la seconde équation : 17l – l² = 60. Soit, sous une formulation davantage propice à la résolution d'équations du second degré : -l² + 17l – 60 = 0. Le discriminant est égal à ? = 289 – (4 × 60) = 49, soit le carré de 7.

Comment résoudre une équation sous forme de problème ?

Calcul du discriminant : ? = b2 ?4ac = ( ?2)2 ?4(1)(1) = ?2.
. Le discriminant est strictement négatif, la règle est donc "toujours du signe de a", c'est à dire toujours positif car a = 1.

Comment résoudre un système d'équation du second degré ?

- Si ? > 0, alors l'équation admet deux solutions réelles notées x1 et x2.
. On a alors : x1 = (?b ? ?? ) / (2a) et x2 = (?b + ?? ) / (2a) ; - Si ? = 0, alors l'équation admet une solution réelle double notée x0.










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